GSEB Class 9 Maths Solutions Chapter 12 Herons Formula Exercise 12.2

Question 1. A park in the shape of a quadrilateral ABCD, has \( \angle C = 90^\circ \), \( AB = 9m \), \( BC = 12m \), \( CD = 5m \) and \( AD = 8 m \). How much area does it occupy?
Answer: First, join the diagonal BD. In the right-angled triangle BCD, we can find the length of BD using the Pythagorean theorem.
\( BD^2 = BC^2 + CD^2 \)
\( BD^2 = (12)^2 + (5)^2 \)
\( BD^2 = 144 + 25 \)
\( BD^2 = 169 \)
\( BD = \sqrt{169} \)
\( BD = 13m \)
The area of the right triangle BCD is calculated as:
Area \( = \frac {1}{2} \times \text{base} \times \text{height} = \frac {1}{2} \times 5 \times 12 = 30m^2 \)
For triangle ABD, the sides are \( a = 13m \), \( b = 8m \), and \( c = 9m \). We use Heron's formula to get its area.
The semi-perimeter \( s = \frac {a + b + c}{2} = \frac {13 + 8 + 9}{2} = \frac {30}{2} = 15m \)
Area of \( \Delta ABD = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{15(15-13)(15-8)(15-9)} \)
\( = \sqrt{15 \times 2 \times 7 \times 6} \)
\( = \sqrt{(3 \times 5) \times 2 \times 7 \times (2 \times 3)} \)
\( = \sqrt{3^2 \times 5 \times 2^2 \times 7} \)
\( = 3 \times 2 \sqrt{5 \times 7} \)
\( = 6 \sqrt{35}m^2 \)
Using the approximate value \( \sqrt{35} \approx 5.916 \):
Area of \( \Delta ABD = 6 \times 5.916 = 35.496 \approx 35.5 m^2 \)
The total area of the quadrilateral ABCD is the sum of the areas of \( \Delta BCD \) and \( \Delta ABD \).
Total Area \( = 30 m^2 + 35.5 m^2 = 65.5 m^2 \)
Therefore, the park occupies an area of approximately \( 65.5 m^2 \).
In simple words: We split the park into two triangles. First, we found the missing side using a right triangle rule. Then, we calculated the area of both triangles and added them together to get the total area.

Exam Tip: Remember to break down complex shapes into simpler ones like triangles. For right-angled triangles, use \( \frac{1}{2} \times \text{base} \times \text{height} \); for other triangles, use Heron's formula.

Question 2. Find the area of a quadrilateral ABCD in which \( AB = 3 \) cm, \( BC = 4 \) cm, \( CD = 4 \) cm, \( DA = 5 \) cm and \( AC = 5 \) cm.
Answer: We can divide the quadrilateral ABCD into two triangles, \( \Delta ABC \) and \( \Delta ACD \), by drawing the diagonal AC.
For \( \Delta ABC \), the sides are \( AB = 3 \) cm, \( BC = 4 \) cm, and \( AC = 5 \) cm.
We check if it is a right-angled triangle using the Pythagorean theorem: \( 3^2 + 4^2 = 9 + 16 = 25 \), which is equal to \( 5^2 \). Since \( AB^2 + BC^2 = AC^2 \), \( \Delta ABC \) is a right-angled triangle with the right angle at B.
Area of \( \Delta ABC = \frac {1}{2} \times \text{base} \times \text{height} = \frac {1}{2} \times AB \times BC = \frac {1}{2} \times 3 \times 4 = 6 \text{ cm}^2 \).
For \( \Delta ACD \), the sides are \( AC = 5 \) cm, \( CD = 4 \) cm, and \( DA = 5 \) cm.
We will use Heron's formula to find its area.
The semi-perimeter \( s = \frac {AC + CD + DA}{2} = \frac {5 + 4 + 5}{2} = \frac {14}{2} = 7 \text{ cm} \).
Area of \( \Delta ACD = \sqrt{s(s-AC)(s-CD)(s-DA)} \)
\( = \sqrt{7(7-5)(7-4)(7-5)} \)
\( = \sqrt{7 \times 2 \times 3 \times 2} \)
\( = \sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21} \text{ cm}^2 \).
Using the approximate value \( \sqrt{21} \approx 4.58 \):
Area of \( \Delta ACD = 2 \times 4.58 \approx 9.16 \text{ cm}^2 \). (The source uses \( 4.6 \) for \( \sqrt{21} \) resulting in \( 9.2 \text{ cm}^2 \)).
The total area of the quadrilateral ABCD is the sum of the areas of \( \Delta ABC \) and \( \Delta ACD \).
Total Area \( = 6 \text{ cm}^2 + 9.2 \text{ cm}^2 = 15.2 \text{ cm}^2 \).
In simple words: We split the quadrilateral into two triangles. One was a right-angled triangle, so we used a simple area formula. For the other, we used Heron's formula. Then we added both areas to get the total area of the quadrilateral.

Exam Tip: Always check if a triangle formed by a diagonal is a right-angled triangle by using the Pythagorean theorem, as it simplifies area calculation. Otherwise, Heron's formula is a reliable method.

Question 3. Radha made a picture of an aeroplane with colored paper as shown in figure. Find the total area of the paper used.
Answer: We need to calculate the area of five different parts of the aeroplane model and then sum them up.
**Area I (Triangular part):**
This is an isosceles triangle with sides \( a = 5 \) cm, \( b = 5 \) cm, and \( c = 1 \) cm.
The semi-perimeter \( s = \frac {5 + 5 + 1}{2} = \frac {11}{2} = 5.5 \text{ cm} \).
Area I \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{5.5(5.5-5)(5.5-5)(5.5-1)} \)
\( = \sqrt{5.5 \times 0.5 \times 0.5 \times 4.5} \)
\( = 0.5 \sqrt{5.5 \times 4.5} \)
\( = 0.5 \sqrt{0.5 \times 11 \times 0.5 \times 9} \)
\( = 0.5 \times 0.5 \times 3 \sqrt{11} \)
\( = 0.75 \sqrt{11} \)
Using \( \sqrt{11} \approx 3.3 \):
Area I \( = 0.75 \times 3.3 \approx 2.475 \approx 2.5 \text{ cm}^2 \).
**Area II (Rectangular part):**
This is a rectangle with length 6.5 cm and width 1 cm.
Area II \( = \text{length} \times \text{width} = 6.5 \times 1 = 6.5 \text{ cm}^2 \).
**Area III (Trapezium part):**
This is an isosceles trapezium with parallel sides 1 cm and 2 cm, and non-parallel sides 1 cm each. To find the height (h), we can form two right triangles by drawing perpendiculars from the ends of the 1 cm side to the 2 cm side. The base of these triangles will be \( \frac{2-1}{2} = 0.5 \) cm.
\( h^2 = 1^2 - (0.5)^2 = 1 - 0.25 = 0.75 \)
\( h = \sqrt{0.75} = \frac{\sqrt{3}}{2} \text{ cm} \).
Area III \( = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \)
\( = \frac{1}{2} \times (1 + 2) \times \frac{\sqrt{3}}{2} \)
\( = \frac{3\sqrt{3}}{4} \text{ cm}^2 \)
Using \( \sqrt{3} \approx 1.732 \):
Area III \( = \frac{3 \times 1.732}{4} = \frac{5.196}{4} \approx 1.3 \text{ cm}^2 \).
**Area IV (Triangular wing):**
This is a right-angled triangle with base 6 cm and height 1.5 cm.
Area IV \( = \frac {1}{2} \times 6 \times 1.5 = 4.5 \text{ cm}^2 \).
**Area V (Triangular wing):**
This is identical to Area IV.
Area V \( = \frac {1}{2} \times 6 \times 1.5 = 4.5 \text{ cm}^2 \).
**Total Area of the paper used:**
Total Area \( = \text{Area I} + \text{Area II} + \text{Area III} + \text{Area IV} + \text{Area V} \)
\( = 2.5 \text{ cm}^2 + 6.5 \text{ cm}^2 + 1.3 \text{ cm}^2 + 4.5 \text{ cm}^2 + 4.5 \text{ cm}^2 \)
\( = 19.3 \text{ cm}^2 \).
In simple words: We separated the aeroplane picture into five parts: one main triangle, one rectangle, one trapezium, and two smaller wing triangles. We calculated the area of each part using suitable formulas and then added all these areas together to find the total paper used.

Exam Tip: For complex figures, break them down into basic geometric shapes. Label each part clearly and calculate its area individually before summing them up for the total area. Be careful with measurements from diagrams.

Question 4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm, and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Answer: First, we need to find the area of the triangle using Heron's formula.
The sides of the triangle are \( a = 26 \) cm, \( b = 28 \) cm, and \( c = 30 \) cm.
The semi-perimeter \( s = \frac {a + b + c}{2} = \frac {26 + 28 + 30}{2} = \frac {84}{2} = 42 \text{ cm} \).
Area of the triangle \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{42(42-26)(42-28)(42-30)} \)
\( = \sqrt{42 \times 16 \times 14 \times 12} \)
\( = \sqrt{(6 \times 7) \times 16 \times (2 \times 7) \times (2 \times 6)} \)
\( = \sqrt{6^2 \times 7^2 \times 16 \times 2^2} \)
\( = 6 \times 7 \times 4 \times 2 \)
\( = 336 \text{ cm}^2 \).
Now, for the parallelogram, its base is given as 28 cm. Let the height of the parallelogram be \( h \) cm.
The area of a parallelogram \( = \text{Base} \times \text{Height} = 28 \times h \text{ cm}^2 \).
According to the question, the area of the triangle is equal to the area of the parallelogram.
So, \( 28h = 336 \)
\( h = \frac{336}{28} \)
\( h = 12 \text{ cm} \).
Therefore, the height of the parallelogram is 12 cm.
In simple words: We first calculated the area of the triangle using its three side lengths. Since the triangle and parallelogram have the same area and the parallelogram's base is known, we used the area of a parallelogram formula (base times height) to find its height.

Exam Tip: This problem links two key geometric formulas: Heron's formula for triangle area and the base-times-height formula for parallelogram area. Make sure you can apply both correctly and understand when to equate areas.

Question 5. A rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Answer: A rhombus is made up of two congruent triangles. We can find the area of one of these triangles and then multiply it by two to get the total area of the rhombus.
Let's consider one triangle, say \( \Delta ABC \). The sides are \( a = 30 \) m, \( b = 48 \) m (the longer diagonal), and \( c = 30 \) m.
First, calculate the semi-perimeter \( s \) for \( \Delta ABC \):
\( s = \frac {30 + 48 + 30}{2} = \frac {108}{2} = 54 \text{ m} \).
Now, use Heron's formula to find the area of \( \Delta ABC \):
Area of \( \Delta ABC = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{54(54-30)(54-48)(54-30)} \)
\( = \sqrt{54 \times 24 \times 6 \times 24} \)
\( = \sqrt{(9 \times 6) \times 24 \times 6 \times 24} \)
\( = \sqrt{9 \times 6^2 \times 24^2} \)
\( = 3 \times 6 \times 24 \)
\( = 18 \times 24 \)
\( = 432 \text{ m}^2 \).
The total area of the rhombus is twice the area of \( \Delta ABC \):
Area of rhombus \( = 2 \times 432 = 864 \text{ m}^2 \).
This total area is for 18 cows to graze.
To find the area of grass field each cow will get, divide the total area by the number of cows:
Area per cow \( = \frac{864}{18} = 48 \text{ m}^2 \).
Therefore, each cow will get \( 48 \text{ m}^2 \) of grass field.
In simple words: We split the rhombus into two identical triangles. We found the area of one triangle using its side lengths and then doubled it for the total rhombus area. Finally, we divided this total area by the number of cows to see how much grass each cow gets.

Exam Tip: Remember that a rhombus can be divided into two congruent triangles by its diagonal. Applying Heron's formula to one triangle and then doubling the result is an efficient way to find the rhombus's area. Don't forget the final step of dividing the total area if the question asks for per-unit allocation.

Question 6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
Answer: First, we need to calculate the area of one triangular piece of cloth. Each piece has sides measuring \( a = 20 \) cm, \( b = 50 \) cm, and \( c = 50 \) cm.
Using Heron's formula, we start by finding the semi-perimeter \( s \):
\( s = \frac {a + b + c}{2} = \frac {20 + 50 + 50}{2} = \frac {120}{2} = 60 \text{ cm} \).
Now, calculate the area of one triangular piece:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{60(60-20)(60-50)(60-50)} \)
\( = \sqrt{60 \times 40 \times 10 \times 10} \)
\( = \sqrt{(6 \times 10) \times (4 \times 10) \times 10 \times 10} \)
\( = \sqrt{6 \times 4 \times 10^4} \)
\( = 100 \sqrt{24} \)
\( = 100 \times \sqrt{4 \times 6} \)
\( = 100 \times 2 \sqrt{6} \)
\( = 200\sqrt{6} \text{ cm}^2 \).
The umbrella uses 10 triangular pieces in total, made of two different colors. This means there are \( \frac{10}{2} = 5 \) pieces of each color.
The total cloth required for one color is the area of 5 such pieces:
Cloth of one color \( = 5 \times 200\sqrt{6} = 1000\sqrt{6} \text{ cm}^2 \).
Therefore, \( 1000\sqrt{6} \text{ cm}^2 \) of cloth of each color is required for the umbrella.
In simple words: We first found the area of one triangle using Heron's formula. Since there are 10 triangles of two colors, each color uses 5 triangles. We multiplied the area of one triangle by 5 to get the total cloth needed for each color.

Exam Tip: When dealing with multiple identical pieces, first calculate the area of a single unit, then multiply by the number of units of that type. Pay attention to details like different colors or materials.

Question 7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and side 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?
Answer: The kite is divided into three shades. Shades I and II form the square part, and Shade III is the isosceles triangle at the bottom.
**Area of Shade I and Shade II:**
The square has a diagonal of 32 cm. In a square, both diagonals are equal and bisect each other at right angles. So, the area of the square is \( \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 32 \times 32 = 512 \text{ cm}^2 \).
The figure shows the square is divided into two equal parts for Shade I and Shade II. Each part is a triangle with base 32 cm (the diagonal) and height \( \frac{32}{2} = 16 \) cm (half of the other diagonal).
Area of Shade I \( = \frac{1}{2} \times 32 \times 16 = 256 \text{ cm}^2 \).
Area of Shade II \( = \frac{1}{2} \times 32 \times 16 = 256 \text{ cm}^2 \).
**Area of Shade III:**
This is an isosceles triangle with a base of 8 cm and other sides of 6 cm each. Let the sides be \( a = 6 \) cm, \( b = 6 \) cm, and \( c = 8 \) cm.
First, find the semi-perimeter \( s \):
\( s = \frac {6 + 6 + 8}{2} = \frac {20}{2} = 10 \text{ cm} \).
Now, use Heron's formula to find the area of Shade III:
Area III \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{10(10-6)(10-6)(10-8)} \)
\( = \sqrt{10 \times 4 \times 4 \times 2} \)
\( = \sqrt{320} = \sqrt{64 \times 5} = 8\sqrt{5} \text{ cm}^2 \).
Using \( \sqrt{5} \approx 2.236 \):
Area III \( = 8 \times 2.236 \approx 17.888 \approx 17.89 \text{ cm}^2 \).
So, the paper used for each shade is:
Paper for Shade I: \( 256 \text{ cm}^2 \)
Paper for Shade II: \( 256 \text{ cm}^2 \)
Paper for Shade III: \( 8\sqrt{5} \text{ cm}^2 \) (approximately \( 17.89 \text{ cm}^2 \)).
In simple words: The kite is split into three parts. The top two parts make up a square, so we found the area of each half of the square. The bottom part is a triangle, and we calculated its area using Heron's formula with its given side lengths.

Exam Tip: Recognize that a square's area can be calculated using its diagonal \( (\frac{1}{2}d^2) \). For a kite, divide it into its component triangles and use appropriate area formulas for each part, then sum up or list them separately as requested.

Question 8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm. Find the cost of polishing the tiles at the rate of 50 paise per cm².
Answer: First, we need to find the area of one triangular tile. The sides of each triangle are \( a = 9 \) cm, \( b = 28 \) cm, and \( c = 35 \) cm.
Using Heron's formula, we calculate the semi-perimeter \( s \):
\( s = \frac {9 + 28 + 35}{2} = \frac {72}{2} = 36 \text{ cm} \).
Now, find the area of one tile:
Area of one tile \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{36(36-9)(36-28)(36-35)} \)
\( = \sqrt{36 \times 27 \times 8 \times 1} \)
\( = \sqrt{36 \times (9 \times 3) \times (4 \times 2)} \)
\( = \sqrt{36 \times 9 \times 4 \times 6} \)
\( = \sqrt{6^2 \times 3^2 \times 2^2 \times 6} \)
\( = 6 \times 3 \times 2 \sqrt{6} \)
\( = 36\sqrt{6} \text{ cm}^2 \).
There are 16 such tiles in the floral design. So, the total area to be polished is:
Total Area \( = 16 \times 36\sqrt{6} = 576\sqrt{6} \text{ cm}^2 \).
The cost of polishing is 50 paise per cm². We convert 50 paise to Rupees: \( 50 \text{ paise} = \text{Rs. } \frac{50}{100} = \text{Rs. } 0.50 \).
Total cost of polishing \( = \text{Total Area} \times \text{Rate per cm}^2 \)
\( = 576\sqrt{6} \times 0.50 \)
\( = 288\sqrt{6} \text{ Rs.} \).
Using \( \sqrt{6} \approx 2.449 \):
Total cost \( = 288 \times 2.449 \approx 705.312 \approx \text{Rs. } 705.60 \).
In simple words: We calculated the area of one triangular tile using Heron's formula. Then, we multiplied this by 16 to find the total area of all the tiles. Finally, we multiplied the total area by the polishing cost per square centimeter to get the final cost.

Exam Tip: Remember to convert paise to Rupees at the end of the calculation to avoid errors. It's often easier to convert the rate first if there are many calculations involved.

Question 9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The nonparallel sides are 14 m and 13 m. Find the area of the field.
Answer: Let the trapezium be ABCD, where AB is parallel to CD. Let \( AB = 25 \) m, \( CD = 10 \) m, \( AD = 14 \) m, and \( BC = 13 \) m.
To find the area, we draw a line DE parallel to BC from D, meeting AB at E.
This forms a parallelogram EBCD (since DE is parallel to BC and CD is parallel to BE).
In parallelogram EBCD:
\( BE = CD = 10 \) m
\( DE = BC = 13 \) m
Now, consider the triangle AED:
\( AE = AB - BE = 25 - 10 = 15 \) m.
So, the sides of \( \Delta AED \) are \( AD = 14 \) m, \( DE = 13 \) m, and \( AE = 15 \) m.
First, we find the area of \( \Delta AED \) using Heron's formula.
The semi-perimeter \( s = \frac {AD + DE + AE}{2} = \frac {14 + 13 + 15}{2} = \frac {42}{2} = 21 \text{ m} \).
Area of \( \Delta AED = \sqrt{s(s-AD)(s-DE)(s-AE)} \)
\( = \sqrt{21(21-14)(21-13)(21-15)} \)
\( = \sqrt{21 \times 7 \times 8 \times 6} \)
\( = \sqrt{(3 \times 7) \times 7 \times (2^3) \times (2 \times 3)} \)
\( = \sqrt{2^4 \times 3^2 \times 7^2} \)
\( = 2^2 \times 3 \times 7 \)
\( = 4 \times 3 \times 7 = 84 \text{ m}^2 \).
Next, we find the height (h) of \( \Delta AED \) corresponding to the base AE. This height will also be the height of the trapezium.
Area of \( \Delta AED = \frac{1}{2} \times \text{base} \times \text{height} \)
\( 84 = \frac{1}{2} \times AE \times h \)
\( 84 = \frac{1}{2} \times 15 \times h \)
\( h = \frac{84 \times 2}{15} = \frac{168}{15} = 11.2 \text{ m} \).
Now, find the area of the parallelogram EBCD:
Area of parallelogram EBCD \( = \text{Base} \times \text{Height} = BE \times h = 10 \times 11.2 = 112 \text{ m}^2 \).
Finally, the total area of the field (trapezium ABCD) is the sum of the areas of \( \Delta AED \) and parallelogram EBCD.
Total Area \( = 84 \text{ m}^2 + 112 \text{ m}^2 = 196 \text{ m}^2 \).
Therefore, the area of the field is \( 196 \text{ m}^2 \).
In simple words: We split the trapezium into a parallelogram and a triangle. We used Heron's formula to find the area of the triangle and then calculated its height. This height is also the height of the parallelogram. We found the area of the parallelogram and added it to the triangle's area to get the total area of the field.

Exam Tip: For trapeziums with non-parallel sides, a common strategy is to construct a parallelogram and a triangle. This allows you to apply Heron's formula for the triangle and then calculate the height needed for both the triangle and the parallelogram.